python – 移动Flask-Restplus Swagger API文档
作者：互联网
我正在尝试使用flask-restplus在python中构建一个restful API.我希望将swagger文档放在与普通“/”不同的地方.
我正在按照文档here并按照说明操作.我正在使用python2.7.3并具有以下代码〜/ dev / test / app.py：
from flask import Flask
from flask.ext.restplus import Api, apidoc
app = Flask(__name__)
api = Api(app, ui=False)
@api.route('/doc/', endpoint='doc')
def swagger_ui():
return apidoc.ui_for(api)
app.register_blueprint(apidoc.apidoc)
当我尝试运行这个python app.py时,我得到：
Traceback (most recent call last):
File "app.py", line 7 in <module>
@api.route('/doc/', endpoint='doc')
File "/home/logan/.virtualenvs/test/lib/python2.7/site-packages/flask_restplus/api.py", line 191, in wrapper
self.add_resources(cls, *urls, **kwargs)
File "/home/logan/.virtualenvs/test/lib/python2.7/site-packages/flask_restplus/api.py", line 175, in add_resource
super(Api, self).add_resource(resource, *urls, **kwargs)
File "/home/logan/.virtualenvs/test/lib/python2.7/site-packages/flask_restful/__init__.py", line 396, in add_resource
self._register_view(self.app, resource, *urls, **kwargs)
File "/home/logan/.virtualenvs/test/lib/python2.7/site-packages/flask_restful/__init__.py", line 435, in _register_view
resource_func = self.output(resource.as_view(endpoint, *resource_class_args,
AttributeError: 'function' object has no attribute 'as_view'
我不确定到底出了什么问题,我想我明白我没有从资源中继承as_view通常来自的地方,但文档似乎表明这应该有效.
任何帮助都会得到帮助.
解决方法:
使用Flask-Restplus< = 0.8.0,你应该写：
from flask import Flask
from flask.ext.restplus import Api, apidoc
app = Flask(__name__)
api = Api(app, ui=False)
@app.route('/doc/', endpoint='doc')
def swagger_ui():
return apidoc.ui_for(api)
注意使用@app而不是@api
从v0.8.1(即将发布)开始,你只需要写：
from flask import Flask
from flask.ext.restplus import Api, apidoc
app = Flask(__name__)
api = Api(app, doc='/doc/')
见：http://flask-restplus.readthedocs.org/en/latest/swagger.html#swagger-ui
标签：swagger-ui,python,swagger,flask,flask-restplus 来源： https://codeday.me/bug/20190824/1709129.html